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Is $f:[0,10]tomathbb Rmid f(x)=x-[x]$ Riemann-Integrable?
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True or false?
The function $$f:[0,10]tomathbb Rmid f(x)=x-[x]$$ is Riemann-Integrable.
Here $[x]$ is the floor function, and I think $x-[x]$ is the mantissa function:
I think the statement is true, but I don't know how to prove in a easy way. Maybe proving that if $f$ is bounded (it's true) then $$underlinedisplaystyleint_a^bf(x);text dxquad=quadoverlinedisplaystyleint_a^bf(x);text dx,,tag 1$$ but this could be a little tedious?
What are the hypothesis, and my thesis is true? We need to check $(1)$?
Any ideas?
Thank you!
proof-writing riemann-integration
asked Aug 1 at 9:10
manooooh
422211
add a comment |Â
up vote
0
down vote
favorite
True or false?
The function $$f:[0,10]tomathbb Rmid f(x)=x-[x]$$ is Riemann-Integrable.
Here $[x]$ is the floor function, and I think $x-[x]$ is the mantissa function:
I think the statement is true, but I don't know how to prove in a easy way. Maybe proving that if $f$ is bounded (it's true) then $$underlinedisplaystyleint_a^bf(x);text dxquad=quadoverlinedisplaystyleint_a^bf(x);text dx,,tag 1$$ but this could be a little tedious?
What are the hypothesis, and my thesis is true? We need to check $(1)$?
Any ideas?
Thank you!
proof-writing riemann-integration
asked Aug 1 at 9:10
manooooh
422211
4
See math.stackexchange.com/questions/102844/…?
– Robert Z
Aug 1 at 9:15
Thanks! I am very new with this type of integrals. I will take a look.
– manooooh
Aug 1 at 9:21
1
The function considered is a bounded, piecewise continuous function on the interval of integration. The details of the particular function are not necessary to prove its integrability.
– Dan Fox
Aug 1 at 9:22
@DanFox what do you mean by "particular function"?
– manooooh
Aug 1 at 9:27
2
@manooooh: I mean it does not matter that you consider $f(x) = x - [x]$. To prove the Riemann integrability, it is enough that the function considered is piecewise continuous and bounded on the interval of integration. In the case of the particular function you consider, it has to be Riemann integrable because just by looking one can calculate the area under its graph ...
– Dan Fox
Aug 1 at 9:29
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
True or false?
The function $$f:[0,10]tomathbb Rmid f(x)=x-[x]$$ is Riemann-Integrable.
Here $[x]$ is the floor function, and I think $x-[x]$ is the mantissa function:
I think the statement is true, but I don't know how to prove in a easy way. Maybe proving that if $f$ is bounded (it's true) then $$underlinedisplaystyleint_a^bf(x);text dxquad=quadoverlinedisplaystyleint_a^bf(x);text dx,,tag 1$$ but this could be a little tedious?
What are the hypothesis, and my thesis is true? We need to check $(1)$?
Any ideas?
Thank you!
proof-writing riemann-integration
asked Aug 1 at 9:10
manooooh
422211
True or false?
The function $$f:[0,10]tomathbb Rmid f(x)=x-[x]$$ is Riemann-Integrable.
Here $[x]$ is the floor function, and I think $x-[x]$ is the mantissa function:
I think the statement is true, but I don't know how to prove in a easy way. Maybe proving that if $f$ is bounded (it's true) then $$underlinedisplaystyleint_a^bf(x);text dxquad=quadoverlinedisplaystyleint_a^bf(x);text dx,,tag 1$$ but this could be a little tedious?
What are the hypothesis, and my thesis is true? We need to check $(1)$?
Any ideas?
Thank you!
proof-writing riemann-integration
asked Aug 1 at 9:10
manooooh
422211
edited Aug 1 at 9:18
asked Aug 1 at 9:10
manooooh
422211
asked Aug 1 at 9:10
manooooh
422211
asked Aug 1 at 9:10
manooooh
422211
422211
4
See math.stackexchange.com/questions/102844/…?
– Robert Z
Aug 1 at 9:15
Thanks! I am very new with this type of integrals. I will take a look.
– manooooh
Aug 1 at 9:21
1
The function considered is a bounded, piecewise continuous function on the interval of integration. The details of the particular function are not necessary to prove its integrability.
– Dan Fox
Aug 1 at 9:22
@DanFox what do you mean by "particular function"?
– manooooh
Aug 1 at 9:27
2
@manooooh: I mean it does not matter that you consider $f(x) = x - [x]$. To prove the Riemann integrability, it is enough that the function considered is piecewise continuous and bounded on the interval of integration. In the case of the particular function you consider, it has to be Riemann integrable because just by looking one can calculate the area under its graph ...
– Dan Fox
Aug 1 at 9:29
add a comment |Â
4
See math.stackexchange.com/questions/102844/…?
– Robert Z
Aug 1 at 9:15
Thanks! I am very new with this type of integrals. I will take a look.
– manooooh
Aug 1 at 9:21
1
The function considered is a bounded, piecewise continuous function on the interval of integration. The details of the particular function are not necessary to prove its integrability.
– Dan Fox
Aug 1 at 9:22
@DanFox what do you mean by "particular function"?
– manooooh
Aug 1 at 9:27
2
@manooooh: I mean it does not matter that you consider $f(x) = x - [x]$. To prove the Riemann integrability, it is enough that the function considered is piecewise continuous and bounded on the interval of integration. In the case of the particular function you consider, it has to be Riemann integrable because just by looking one can calculate the area under its graph ...
– Dan Fox
Aug 1 at 9:29
4
4
See math.stackexchange.com/questions/102844/…?
– Robert Z
Aug 1 at 9:15
See math.stackexchange.com/questions/102844/…?
– Robert Z
Aug 1 at 9:15
Thanks! I am very new with this type of integrals. I will take a look.
– manooooh
Aug 1 at 9:21
Thanks! I am very new with this type of integrals. I will take a look.
– manooooh
Aug 1 at 9:21
1
1
The function considered is a bounded, piecewise continuous function on the interval of integration. The details of the particular function are not necessary to prove its integrability.
– Dan Fox
Aug 1 at 9:22
The function considered is a bounded, piecewise continuous function on the interval of integration. The details of the particular function are not necessary to prove its integrability.
– Dan Fox
Aug 1 at 9:22
@DanFox what do you mean by "particular function"?
– manooooh
Aug 1 at 9:27
@DanFox what do you mean by "particular function"?
– manooooh
Aug 1 at 9:27
2
2
@manooooh: I mean it does not matter that you consider $f(x) = x - [x]$. To prove the Riemann integrability, it is enough that the function considered is piecewise continuous and bounded on the interval of integration. In the case of the particular function you consider, it has to be Riemann integrable because just by looking one can calculate the area under its graph ...
– Dan Fox
Aug 1 at 9:29
@manooooh: I mean it does not matter that you consider $f(x) = x - [x]$. To prove the Riemann integrability, it is enough that the function considered is piecewise continuous and bounded on the interval of integration. In the case of the particular function you consider, it has to be Riemann integrable because just by looking one can calculate the area under its graph ...
– Dan Fox
Aug 1 at 9:29
add a comment |Â
1 Answer
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oldest
votes
up vote
1
down vote
accepted
If you want to prove this by way of
$$
underlinedisplaystyleint_a^bf(x);text dxquad=quadoverlinedisplaystyleint_a^bf(x);text dx,,tag 1
$$
then it can be done. Take any $epsilon>0$. We want to make a partition $P = (x_0, x_1, ldots,x_n)$ of $[0,10]$ (with $x_0 = 0, x_n = 10$, and $x_i<x_i+1$) such that
$$
U(f, P) - L(f, P)<epsilon
$$
where $U(f, P)$ is the upper sum of $f$ with respect to the partition $P$ and $L(f, P)$ is the lower. It's not too hard to show that any partition where $x_i+1-x_i < epsilon/20$ will work.
answered Aug 1 at 10:12
Arthur
98.3k793174
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If you want to prove this by way of
$$
underlinedisplaystyleint_a^bf(x);text dxquad=quadoverlinedisplaystyleint_a^bf(x);text dx,,tag 1
$$
then it can be done. Take any $epsilon>0$. We want to make a partition $P = (x_0, x_1, ldots,x_n)$ of $[0,10]$ (with $x_0 = 0, x_n = 10$, and $x_i<x_i+1$) such that
$$
U(f, P) - L(f, P)<epsilon
$$
where $U(f, P)$ is the upper sum of $f$ with respect to the partition $P$ and $L(f, P)$ is the lower. It's not too hard to show that any partition where $x_i+1-x_i < epsilon/20$ will work.
answered Aug 1 at 10:12
Arthur
98.3k793174
add a comment |Â
up vote
1
down vote
accepted
If you want to prove this by way of
$$
underlinedisplaystyleint_a^bf(x);text dxquad=quadoverlinedisplaystyleint_a^bf(x);text dx,,tag 1
$$
then it can be done. Take any $epsilon>0$. We want to make a partition $P = (x_0, x_1, ldots,x_n)$ of $[0,10]$ (with $x_0 = 0, x_n = 10$, and $x_i<x_i+1$) such that
$$
U(f, P) - L(f, P)<epsilon
$$
where $U(f, P)$ is the upper sum of $f$ with respect to the partition $P$ and $L(f, P)$ is the lower. It's not too hard to show that any partition where $x_i+1-x_i < epsilon/20$ will work.
answered Aug 1 at 10:12
Arthur
98.3k793174
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If you want to prove this by way of
$$
underlinedisplaystyleint_a^bf(x);text dxquad=quadoverlinedisplaystyleint_a^bf(x);text dx,,tag 1
$$
then it can be done. Take any $epsilon>0$. We want to make a partition $P = (x_0, x_1, ldots,x_n)$ of $[0,10]$ (with $x_0 = 0, x_n = 10$, and $x_i<x_i+1$) such that
$$
U(f, P) - L(f, P)<epsilon
$$
where $U(f, P)$ is the upper sum of $f$ with respect to the partition $P$ and $L(f, P)$ is the lower. It's not too hard to show that any partition where $x_i+1-x_i < epsilon/20$ will work.
answered Aug 1 at 10:12
Arthur
98.3k793174
If you want to prove this by way of
$$
underlinedisplaystyleint_a^bf(x);text dxquad=quadoverlinedisplaystyleint_a^bf(x);text dx,,tag 1
$$
then it can be done. Take any $epsilon>0$. We want to make a partition $P = (x_0, x_1, ldots,x_n)$ of $[0,10]$ (with $x_0 = 0, x_n = 10$, and $x_i<x_i+1$) such that
$$
U(f, P) - L(f, P)<epsilon
$$
where $U(f, P)$ is the upper sum of $f$ with respect to the partition $P$ and $L(f, P)$ is the lower. It's not too hard to show that any partition where $x_i+1-x_i < epsilon/20$ will work.
answered Aug 1 at 10:12
Arthur
98.3k793174
answered Aug 1 at 10:12
Arthur
98.3k793174
answered Aug 1 at 10:12
Arthur
98.3k793174
answered Aug 1 at 10:12
Arthur
98.3k793174
98.3k793174
add a comment |Â
add a comment |Â
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4
See math.stackexchange.com/questions/102844/…?
– Robert Z
Aug 1 at 9:15
Thanks! I am very new with this type of integrals. I will take a look.
– manooooh
Aug 1 at 9:21
1
The function considered is a bounded, piecewise continuous function on the interval of integration. The details of the particular function are not necessary to prove its integrability.
– Dan Fox
Aug 1 at 9:22
@DanFox what do you mean by "particular function"?
– manooooh
Aug 1 at 9:27
2
@manooooh: I mean it does not matter that you consider $f(x) = x - [x]$. To prove the Riemann integrability, it is enough that the function considered is piecewise continuous and bounded on the interval of integration. In the case of the particular function you consider, it has to be Riemann integrable because just by looking one can calculate the area under its graph ...
– Dan Fox
Aug 1 at 9:29